Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 2}{x + 1} = \dfrac{-5x - 6}{x + 1}$
Answer: Multiply both sides by $x + 1$ $ \dfrac{x^2 - 2}{x + 1} (x + 1) = \dfrac{-5x - 6}{x + 1} (x + 1)$ $ x^2 - 2 = -5x - 6$ Subtract $-5x - 6$ from both sides: $ x^2 - 2 - (-5x - 6) = -5x - 6 - (-5x - 6)$ $ x^2 - 2 + 5x + 6 = 0$ $ x^2 + 4 + 5x = 0$ Factor the expression: $ (x + 1)(x + 4) = 0$ Therefore $x = -1$ or $x = -4$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.